Direct passage from $n$ prime to $n$ non-prime in Euler Totient Function $Phi$












0












$begingroup$


I wish to derive, for a proof, the Euler Totient Function starting from the case $n$ prime to $n$ non-prime.



Let $n$ prime, we know $Phi(n)=n-1$. But what if now I assume $n=p_0^{a_0}p_1^{a_1}...p_r^{a_r}$ and I want to reach $Phi(n)=n prod_{p|n}^{}(1-1/p) $ from there? Like:



$Phi(p_0^{a_0}p_1^{a_1}...p_r^{a_r})=p_0^{a_0}p_1^{a_1}...p_r^{a_r}-1 Rightarrow p_0^{a_0}p_1^{a_1}...p_r^{a_r}(1-1/(p_0^{a_0}p_1^{a_1}...p_r^{a_r}))=?$



Is this kind of approach possible for this formula?






abstract-algebra elementary-number-theory modular-arithmetic totient-function







share|cite|improve this question









$endgroup$








  • 3




    $begingroup$
    The two steps are $phi(p^k) = p^k-p^{k-1}$ and $phi(nm) = phi(n) phi(m)$ whenever $gcd(n,m)=1$ thus $phi(n) = prod_{p^k |n} phi(p^k) = nprod_{p^k |n} frac{phi(p^k) }{p^k}= n prod_{p| n} (1-p^{-1})$
    $endgroup$
    – reuns
    Jan 15 at 8:19








  • 1




    $begingroup$
    In particular, it is not true that $phi(p_0^{a_0}p_1^{a_1}cdots p_r^{a_r})=p_0^{a_0}p_1^{a_1}cdots p_r^{a_r}-1$ when $p_0^{a_0}p_1^{a_1}cdots p_r^{a_r}$ is not prime.
    $endgroup$
    – Greg Martin
    Jan 15 at 8:28










  • $begingroup$
    So such kind of derivation, as I suppose, it's not possible. Thanks
    $endgroup$
    – Alessar
    Jan 15 at 8:31










  • $begingroup$
    @reuns the symbol "$||$" means for "that divide"?
    $endgroup$
    – Alessar
    Jan 15 at 9:13






  • 1




    $begingroup$
    $p^k| n$ iff $p^k | n$ and $p^{k+1} nmid n$. In this setting $p$ means (ranging over) the primes and $p^k$ means over the prime powers
    $endgroup$
    – reuns
    Jan 15 at 9:17


















0












$begingroup$


I wish to derive, for a proof, the Euler Totient Function starting from the case $n$ prime to $n$ non-prime.



Let $n$ prime, we know $Phi(n)=n-1$. But what if now I assume $n=p_0^{a_0}p_1^{a_1}...p_r^{a_r}$ and I want to reach $Phi(n)=n prod_{p|n}^{}(1-1/p) $ from there? Like:



$Phi(p_0^{a_0}p_1^{a_1}...p_r^{a_r})=p_0^{a_0}p_1^{a_1}...p_r^{a_r}-1 Rightarrow p_0^{a_0}p_1^{a_1}...p_r^{a_r}(1-1/(p_0^{a_0}p_1^{a_1}...p_r^{a_r}))=?$



Is this kind of approach possible for this formula?






abstract-algebra elementary-number-theory modular-arithmetic totient-function







share|cite|improve this question









$endgroup$








  • 3




    $begingroup$
    The two steps are $phi(p^k) = p^k-p^{k-1}$ and $phi(nm) = phi(n) phi(m)$ whenever $gcd(n,m)=1$ thus $phi(n) = prod_{p^k |n} phi(p^k) = nprod_{p^k |n} frac{phi(p^k) }{p^k}= n prod_{p| n} (1-p^{-1})$
    $endgroup$
    – reuns
    Jan 15 at 8:19








  • 1




    $begingroup$
    In particular, it is not true that $phi(p_0^{a_0}p_1^{a_1}cdots p_r^{a_r})=p_0^{a_0}p_1^{a_1}cdots p_r^{a_r}-1$ when $p_0^{a_0}p_1^{a_1}cdots p_r^{a_r}$ is not prime.
    $endgroup$
    – Greg Martin
    Jan 15 at 8:28










  • $begingroup$
    So such kind of derivation, as I suppose, it's not possible. Thanks
    $endgroup$
    – Alessar
    Jan 15 at 8:31










  • $begingroup$
    @reuns the symbol "$||$" means for "that divide"?
    $endgroup$
    – Alessar
    Jan 15 at 9:13






  • 1




    $begingroup$
    $p^k| n$ iff $p^k | n$ and $p^{k+1} nmid n$. In this setting $p$ means (ranging over) the primes and $p^k$ means over the prime powers
    $endgroup$
    – reuns
    Jan 15 at 9:17
















0












0








0





$begingroup$


I wish to derive, for a proof, the Euler Totient Function starting from the case $n$ prime to $n$ non-prime.



Let $n$ prime, we know $Phi(n)=n-1$. But what if now I assume $n=p_0^{a_0}p_1^{a_1}...p_r^{a_r}$ and I want to reach $Phi(n)=n prod_{p|n}^{}(1-1/p) $ from there? Like:



$Phi(p_0^{a_0}p_1^{a_1}...p_r^{a_r})=p_0^{a_0}p_1^{a_1}...p_r^{a_r}-1 Rightarrow p_0^{a_0}p_1^{a_1}...p_r^{a_r}(1-1/(p_0^{a_0}p_1^{a_1}...p_r^{a_r}))=?$



Is this kind of approach possible for this formula?






abstract-algebra elementary-number-theory modular-arithmetic totient-function







share|cite|improve this question









$endgroup$




I wish to derive, for a proof, the Euler Totient Function starting from the case $n$ prime to $n$ non-prime.



Let $n$ prime, we know $Phi(n)=n-1$. But what if now I assume $n=p_0^{a_0}p_1^{a_1}...p_r^{a_r}$ and I want to reach $Phi(n)=n prod_{p|n}^{}(1-1/p) $ from there? Like:



$Phi(p_0^{a_0}p_1^{a_1}...p_r^{a_r})=p_0^{a_0}p_1^{a_1}...p_r^{a_r}-1 Rightarrow p_0^{a_0}p_1^{a_1}...p_r^{a_r}(1-1/(p_0^{a_0}p_1^{a_1}...p_r^{a_r}))=?$



Is this kind of approach possible for this formula?






abstract-algebra elementary-number-theory modular-arithmetic totient-function




abstract-algebra elementary-number-theory modular-arithmetic totient-function






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 15 at 8:17









AlessarAlessar

29615




29615








  • 3




    $begingroup$
    The two steps are $phi(p^k) = p^k-p^{k-1}$ and $phi(nm) = phi(n) phi(m)$ whenever $gcd(n,m)=1$ thus $phi(n) = prod_{p^k |n} phi(p^k) = nprod_{p^k |n} frac{phi(p^k) }{p^k}= n prod_{p| n} (1-p^{-1})$
    $endgroup$
    – reuns
    Jan 15 at 8:19








  • 1




    $begingroup$
    In particular, it is not true that $phi(p_0^{a_0}p_1^{a_1}cdots p_r^{a_r})=p_0^{a_0}p_1^{a_1}cdots p_r^{a_r}-1$ when $p_0^{a_0}p_1^{a_1}cdots p_r^{a_r}$ is not prime.
    $endgroup$
    – Greg Martin
    Jan 15 at 8:28










  • $begingroup$
    So such kind of derivation, as I suppose, it's not possible. Thanks
    $endgroup$
    – Alessar
    Jan 15 at 8:31










  • $begingroup$
    @reuns the symbol "$||$" means for "that divide"?
    $endgroup$
    – Alessar
    Jan 15 at 9:13






  • 1




    $begingroup$
    $p^k| n$ iff $p^k | n$ and $p^{k+1} nmid n$. In this setting $p$ means (ranging over) the primes and $p^k$ means over the prime powers
    $endgroup$
    – reuns
    Jan 15 at 9:17
















  • 3




    $begingroup$
    The two steps are $phi(p^k) = p^k-p^{k-1}$ and $phi(nm) = phi(n) phi(m)$ whenever $gcd(n,m)=1$ thus $phi(n) = prod_{p^k |n} phi(p^k) = nprod_{p^k |n} frac{phi(p^k) }{p^k}= n prod_{p| n} (1-p^{-1})$
    $endgroup$
    – reuns
    Jan 15 at 8:19








  • 1




    $begingroup$
    In particular, it is not true that $phi(p_0^{a_0}p_1^{a_1}cdots p_r^{a_r})=p_0^{a_0}p_1^{a_1}cdots p_r^{a_r}-1$ when $p_0^{a_0}p_1^{a_1}cdots p_r^{a_r}$ is not prime.
    $endgroup$
    – Greg Martin
    Jan 15 at 8:28










  • $begingroup$
    So such kind of derivation, as I suppose, it's not possible. Thanks
    $endgroup$
    – Alessar
    Jan 15 at 8:31










  • $begingroup$
    @reuns the symbol "$||$" means for "that divide"?
    $endgroup$
    – Alessar
    Jan 15 at 9:13






  • 1




    $begingroup$
    $p^k| n$ iff $p^k | n$ and $p^{k+1} nmid n$. In this setting $p$ means (ranging over) the primes and $p^k$ means over the prime powers
    $endgroup$
    – reuns
    Jan 15 at 9:17










3




3




$begingroup$
The two steps are $phi(p^k) = p^k-p^{k-1}$ and $phi(nm) = phi(n) phi(m)$ whenever $gcd(n,m)=1$ thus $phi(n) = prod_{p^k |n} phi(p^k) = nprod_{p^k |n} frac{phi(p^k) }{p^k}= n prod_{p| n} (1-p^{-1})$
$endgroup$
– reuns
Jan 15 at 8:19






$begingroup$
The two steps are $phi(p^k) = p^k-p^{k-1}$ and $phi(nm) = phi(n) phi(m)$ whenever $gcd(n,m)=1$ thus $phi(n) = prod_{p^k |n} phi(p^k) = nprod_{p^k |n} frac{phi(p^k) }{p^k}= n prod_{p| n} (1-p^{-1})$
$endgroup$
– reuns
Jan 15 at 8:19






1




1




$begingroup$
In particular, it is not true that $phi(p_0^{a_0}p_1^{a_1}cdots p_r^{a_r})=p_0^{a_0}p_1^{a_1}cdots p_r^{a_r}-1$ when $p_0^{a_0}p_1^{a_1}cdots p_r^{a_r}$ is not prime.
$endgroup$
– Greg Martin
Jan 15 at 8:28




$begingroup$
In particular, it is not true that $phi(p_0^{a_0}p_1^{a_1}cdots p_r^{a_r})=p_0^{a_0}p_1^{a_1}cdots p_r^{a_r}-1$ when $p_0^{a_0}p_1^{a_1}cdots p_r^{a_r}$ is not prime.
$endgroup$
– Greg Martin
Jan 15 at 8:28












$begingroup$
So such kind of derivation, as I suppose, it's not possible. Thanks
$endgroup$
– Alessar
Jan 15 at 8:31




$begingroup$
So such kind of derivation, as I suppose, it's not possible. Thanks
$endgroup$
– Alessar
Jan 15 at 8:31












$begingroup$
@reuns the symbol "$||$" means for "that divide"?
$endgroup$
– Alessar
Jan 15 at 9:13




$begingroup$
@reuns the symbol "$||$" means for "that divide"?
$endgroup$
– Alessar
Jan 15 at 9:13




1




1




$begingroup$
$p^k| n$ iff $p^k | n$ and $p^{k+1} nmid n$. In this setting $p$ means (ranging over) the primes and $p^k$ means over the prime powers
$endgroup$
– reuns
Jan 15 at 9:17






$begingroup$
$p^k| n$ iff $p^k | n$ and $p^{k+1} nmid n$. In this setting $p$ means (ranging over) the primes and $p^k$ means over the prime powers
$endgroup$
– reuns
Jan 15 at 9:17












1 Answer
1






active

oldest

votes


















1












$begingroup$

To be precise, your "kind" of derivation is not possible. This is because $phi(p_0^{a_0}p_1^{a_1}...p_r^{a_r})neq p_0^{a_0}p_1^{a_1}...p_r^{a_r}-1$.



Think about what the totient function entails: the number of integers less than the input that are coprime with it. Only with prime inputs is this one less than the input.



The totient function, however, does retain two properties, namely as reuns mentioned, that $phi(p^k)=p^k-p^{k-1}$ and $phi(n)phi(m)=phi(mn)$ when $m$ and $n$ are coprime. The proof of these is easily verifiable given a google search.



Thus, we can have a "direct" approach. Since powers of primes are always coprime to one another,
begin{align*}
phi(n)&=phi(p_0^{a_0}p_1^{a_1}...p_r^{a_r})\
&=phi(p_0^{a_0})phi(p_1^{a_1})...phi(p_r^{a_r})\
&=(p_0^{a_0}-p_0^{a_0-1})(p_1^{a_1}-p_1^{a_1-1})...(p_r^{a_r}-p_r^{a_r-1})\
&=p_0^{a_0}(1-1/p_0)p_1^{a_1}(1-1/p_1)...p_r^{a_r}(1-1/p_r)\
&=ndisplaystyleprod_{p|n}(1-1/p).
end{align*}






share|cite|improve this answer









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    $begingroup$

    To be precise, your "kind" of derivation is not possible. This is because $phi(p_0^{a_0}p_1^{a_1}...p_r^{a_r})neq p_0^{a_0}p_1^{a_1}...p_r^{a_r}-1$.



    Think about what the totient function entails: the number of integers less than the input that are coprime with it. Only with prime inputs is this one less than the input.



    The totient function, however, does retain two properties, namely as reuns mentioned, that $phi(p^k)=p^k-p^{k-1}$ and $phi(n)phi(m)=phi(mn)$ when $m$ and $n$ are coprime. The proof of these is easily verifiable given a google search.



    Thus, we can have a "direct" approach. Since powers of primes are always coprime to one another,
    begin{align*}
    phi(n)&=phi(p_0^{a_0}p_1^{a_1}...p_r^{a_r})\
    &=phi(p_0^{a_0})phi(p_1^{a_1})...phi(p_r^{a_r})\
    &=(p_0^{a_0}-p_0^{a_0-1})(p_1^{a_1}-p_1^{a_1-1})...(p_r^{a_r}-p_r^{a_r-1})\
    &=p_0^{a_0}(1-1/p_0)p_1^{a_1}(1-1/p_1)...p_r^{a_r}(1-1/p_r)\
    &=ndisplaystyleprod_{p|n}(1-1/p).
    end{align*}






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      To be precise, your "kind" of derivation is not possible. This is because $phi(p_0^{a_0}p_1^{a_1}...p_r^{a_r})neq p_0^{a_0}p_1^{a_1}...p_r^{a_r}-1$.



      Think about what the totient function entails: the number of integers less than the input that are coprime with it. Only with prime inputs is this one less than the input.



      The totient function, however, does retain two properties, namely as reuns mentioned, that $phi(p^k)=p^k-p^{k-1}$ and $phi(n)phi(m)=phi(mn)$ when $m$ and $n$ are coprime. The proof of these is easily verifiable given a google search.



      Thus, we can have a "direct" approach. Since powers of primes are always coprime to one another,
      begin{align*}
      phi(n)&=phi(p_0^{a_0}p_1^{a_1}...p_r^{a_r})\
      &=phi(p_0^{a_0})phi(p_1^{a_1})...phi(p_r^{a_r})\
      &=(p_0^{a_0}-p_0^{a_0-1})(p_1^{a_1}-p_1^{a_1-1})...(p_r^{a_r}-p_r^{a_r-1})\
      &=p_0^{a_0}(1-1/p_0)p_1^{a_1}(1-1/p_1)...p_r^{a_r}(1-1/p_r)\
      &=ndisplaystyleprod_{p|n}(1-1/p).
      end{align*}






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        To be precise, your "kind" of derivation is not possible. This is because $phi(p_0^{a_0}p_1^{a_1}...p_r^{a_r})neq p_0^{a_0}p_1^{a_1}...p_r^{a_r}-1$.



        Think about what the totient function entails: the number of integers less than the input that are coprime with it. Only with prime inputs is this one less than the input.



        The totient function, however, does retain two properties, namely as reuns mentioned, that $phi(p^k)=p^k-p^{k-1}$ and $phi(n)phi(m)=phi(mn)$ when $m$ and $n$ are coprime. The proof of these is easily verifiable given a google search.



        Thus, we can have a "direct" approach. Since powers of primes are always coprime to one another,
        begin{align*}
        phi(n)&=phi(p_0^{a_0}p_1^{a_1}...p_r^{a_r})\
        &=phi(p_0^{a_0})phi(p_1^{a_1})...phi(p_r^{a_r})\
        &=(p_0^{a_0}-p_0^{a_0-1})(p_1^{a_1}-p_1^{a_1-1})...(p_r^{a_r}-p_r^{a_r-1})\
        &=p_0^{a_0}(1-1/p_0)p_1^{a_1}(1-1/p_1)...p_r^{a_r}(1-1/p_r)\
        &=ndisplaystyleprod_{p|n}(1-1/p).
        end{align*}






        share|cite|improve this answer









        $endgroup$



        To be precise, your "kind" of derivation is not possible. This is because $phi(p_0^{a_0}p_1^{a_1}...p_r^{a_r})neq p_0^{a_0}p_1^{a_1}...p_r^{a_r}-1$.



        Think about what the totient function entails: the number of integers less than the input that are coprime with it. Only with prime inputs is this one less than the input.



        The totient function, however, does retain two properties, namely as reuns mentioned, that $phi(p^k)=p^k-p^{k-1}$ and $phi(n)phi(m)=phi(mn)$ when $m$ and $n$ are coprime. The proof of these is easily verifiable given a google search.



        Thus, we can have a "direct" approach. Since powers of primes are always coprime to one another,
        begin{align*}
        phi(n)&=phi(p_0^{a_0}p_1^{a_1}...p_r^{a_r})\
        &=phi(p_0^{a_0})phi(p_1^{a_1})...phi(p_r^{a_r})\
        &=(p_0^{a_0}-p_0^{a_0-1})(p_1^{a_1}-p_1^{a_1-1})...(p_r^{a_r}-p_r^{a_r-1})\
        &=p_0^{a_0}(1-1/p_0)p_1^{a_1}(1-1/p_1)...p_r^{a_r}(1-1/p_r)\
        &=ndisplaystyleprod_{p|n}(1-1/p).
        end{align*}







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 15 at 20:50









        Tejas RaoTejas Rao

        16611




        16611






























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            Antonio Litta Visconti Arese

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            Antonio Litta Visconti Arese I duca Litta, VII marchese di Gambolò In carica 1797 – 1820 Predecessore Pompeo Litta Visconti Arese Successore Alfonso Litta Visconti Arese Trattamento Sua grazia Altri titoli Duca Litta Conte del Regno napoleonico d'Italia Patrizio milanese Don Nascita Milano, 2 maggio 1748 Morte Milano, 14 agosto 1820 Dinastia Litta Visconti Arese Padre Pompeo Litta Visconti Arese Madre Maria Elisabetta Visconti Borromeo Consorte Barbara Barbiano di Belgioioso Religione cattolicesimo Antonio Litta Visconti Arese , I duca Litta, VII marchese di Gambolò (Milano, 2 maggio 1748 – Milano, 14 agosto 1820), è stato un nobile e politico italiano. Composizione "Il pastore svizzero" di Pietro Morlacchi, con dedica Indice 1 Biografia 2 Onorificenze 3 Araldica 4 Albero genealogico 5 Note 6 Altri progetti Biografia | Antonio Litta Visco
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